Question

21. a wire when bent ​​in the form of an equilateral triangle encloses an area of 36​√3 cm². find the area enclosed by the same wire when bent to form:
(i) a square, and
(ii) a rectangle whose length is 2 cm more than its width.

Answer 1

$$\begin{gathered} DearStudent, \\ Letthesidesoftheequilateraltriangleis\text{'}a\text{'} \\ thenweknowthatareaofequilateraltriangleis\frac{\sqrt{3}}{4}{a}^2 \\ Therefore,\frac{\sqrt{3}}{4}{a}^2=36\sqrt{3} \\ \Rightarrow a^2=36\times 4 \\ \Rightarrow a=12cm \\ Hencetotallengthofthewireistheperimeteroftriangle3a=3\times 12=36cm \\ i)whenbenttosquaretheperimeterwillbe36cm \\ Therefore,eachsidesofsquare36/4=9cm \\ Therefore,Areaofsquare=9^2=(sides{)}^2 \\ =81c{m}^2 \\ ii)\sin celengthofrec\tan gleis2cmmorethanwidth \\ Therefore, \\ Letwidth=x \\ thenlength=x+2 \\ perimeter=2[x+(x+2)] \\ =2[2x+2] \\ =4x+4 \\ Butperimeter=36cm \\ Therefore,4x+4=36 \\ x=8cm=width \\ x+2=8+2=10cm=length \\ Therefore,Area=length\times width \\ =10\times 8 \\ =80c{m}^2 \\ Regards. \end{gathered}$$